一.解法

https://leetcode-cn.com/problems/remove-duplicates-from-sorted-array/
要點：雙指針
不要用新數(shù)組存台丛。
Python寒匙，C++鹊奖，Java都用雙指針法印蓖，i表示當(dāng)前位置势似，j去找第一個不等同nums[i]的位置，i++分瘦，注意到最后不要越界需要加一句判斷語句

二.Python實現(xiàn)

class Solution:
    def removeDuplicates(self, nums: List[int]) -> int:
        answer=1
        
        if(len(nums)==0):
            return 0
        if(len(nums)==1):
            return 1

        i=0
        j=0
        
        while(j<len(nums)-1):
            while(nums[j]==nums[i]):
                j+=1
                if(j>len(nums)-1):
                     break;
            
            if(j>len(nums)-1):
                 break
            nums[i+1]=nums[j]
            answer+=1
            i+=1
        
        return answer

三.C++實現(xiàn)

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        int answer=1;
        int i,j;
        if(nums.size()==0) return 0;
        if(nums.size()==1) return 1;

        i=0;
        j=0;
        
        while(j<nums.size()-1){
            while(nums[j]==nums[i]){
                j++;
                if(j>nums.size()-1) break;
            }
            if(j>nums.size()-1) break;
            nums[i+1]=nums[j];
            answer++;
            i++;
        }
    
        return answer;
    }
};

四.java實現(xiàn)

class Solution {
    public int removeDuplicates(int[] nums) {
        int answer=1;
        int i,j;
        if(nums.length==0) return 0;
        if(nums.length==1) return 1;

        i=0;
        j=0;
        
        while(j<nums.length-1){
            while(nums[j]==nums[i]){
                j++;
                if(j>nums.length-1) break;
            }
            if(j>nums.length-1) break;
            nums[i+1]=nums[j];
            answer++;
            i++;
        }
    
        return answer;

    }
}